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Difference between revisions of "2014 AIME II Problems/Problem 5"

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==Solution==
 
==Solution==
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Let r, s, -r-s be the roots of p(x) (per Vieta's). Then <math>r^3 + ar + b = 0</math> and similarly for s. Also, <cmath>q(r+4) = (r+4)^3 + a(r+4) + b = 12r^2 + 48r + 64 + 4a = 0</cmath> Set up a similar equation for s: <cmath>q(s-3) = (s-3)^3 + a(s-3) + b + 240 = -9s^2 + 27s + 213 - 3a = 0</cmath>. Simplifying and adding the equations gives <cmath>3r^2 - 3s^2 + 12r + 9s + 87 = 0 (*)</cmath> Now, let's deal with the a*x. Equating the a in both equations (per Vieta) <cmath>rs + (-r-s)(r+s) = (r+4)(s-3) + (-r-s-1)(r+s+1)</cmath>, which eventually simplifies to <cmath>s = \frac{13 + 5r}{2}</cmath> Substitution into (*) should give r = -5 and r = 1, corresponding to s = -6 and s = 9, and |b| = 330 and 90, for an answer of <math>\boxed{420}</math>.
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Let r, s, -r-s be the roots of p(x) (per Vieta's). Then <math>r^3 + ar + b = 0</math> and similarly for s. Also, <cmath>q(r+4) = (r+4)^3 + a(r+4) + b + 240 = 12r^2 + 48r + 304 + 4a = 0</cmath> Set up a similar equation for s: <cmath>q(s-3) = (s-3)^3 + a(s-3) + b + 240 = -9s^2 + 27s + 213 - 3a = 0</cmath>. Simplifying and adding the equations gives <cmath>3r^2 - 3s^2 + 12r + 9s + 147 = 0 (*)</cmath> Now, let's deal with the a*x. Equating the a in both equations (per Vieta) <cmath>rs + (-r-s)(r+s) = (r+4)(s-3) + (-r-s-1)(r+s+1)</cmath>, which eventually simplifies to <cmath>s = \frac{13 + 5r}{2}</cmath> Substitution into (*) should give r = -5 and r = 1, corresponding to s = -6 and s = 9, and |b| = 330 and 90, for an answer of <math>\boxed{420}</math>.

Revision as of 11:40, 31 March 2014

Problem 5

Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$, and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$. Find the sum of all possible values of $|b|$.


Solution

Let r, s, -r-s be the roots of p(x) (per Vieta's). Then $r^3 + ar + b = 0$ and similarly for s. Also, \[q(r+4) = (r+4)^3 + a(r+4) + b + 240 = 12r^2 + 48r + 304 + 4a = 0\] Set up a similar equation for s: \[q(s-3) = (s-3)^3 + a(s-3) + b + 240 = -9s^2 + 27s + 213 - 3a = 0\]. Simplifying and adding the equations gives \[3r^2 - 3s^2 + 12r + 9s + 147 = 0 (*)\] Now, let's deal with the a*x. Equating the a in both equations (per Vieta) \[rs + (-r-s)(r+s) = (r+4)(s-3) + (-r-s-1)(r+s+1)\], which eventually simplifies to \[s = \frac{13 + 5r}{2}\] Substitution into (*) should give r = -5 and r = 1, corresponding to s = -6 and s = 9, and |b| = 330 and 90, for an answer of $\boxed{420}$.

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